Q113 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics
$\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x=$
- Correct answerA$(a+x) \tan ^{-1} \sqrt{\frac{x}{a}}-\sqrt{a x}+c$
- B$\frac{1}{a+x} \tan ^{-1}\left(\frac{x}{a}\right)-\sqrt{a x}+c$
- C$(a+x) \tan ^{-1}\left(\frac{a}{x}\right)+\sqrt{a x}+c$
- D$\sqrt{a+x} \tan ^{-1}\left(\frac{x}{a}\right)+a x+c$
Solution
$\mathrm{I}=\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$
$\begin{aligned} & \text { Let } x=a \tan ^2 t \Rightarrow d x=2 a \tan t \sec ^2 t d t \\ & \mathrm{I}=\int \sin ^{-1} \sqrt{\frac{a \tan ^2 t}{a+a \tan ^2 t}}\left(2 a \tan t \sec ^2 t\right) d t \\ & =2 a \int \sin ^{-1} \sqrt{\frac{a \tan ^2 t}{a \sec ^2 t}} \tan t \sec ^2 t d t \\ & =2 a \int t\left(\tan \sec ^2 t\right) d t\end{aligned}$
$\begin{aligned} & =2 a\left[t \int \tan t \sec ^2 t d t-\int \frac{d t}{d t} \int \tan t \sec ^2 t d t\right] \\ & =2 a\left[\frac{t \tan ^2 t}{2}-\int \frac{\tan ^2 t}{2} d t\right] \\ & =a t \tan ^2 t-a \int\left(\sec ^2 t-1\right) d t \\ & =\text { at } \tan ^2 t-a \tan t+\mathrm{at}+\mathrm{C} \\ & \text { Put } t=\tan ^{-1} \sqrt{\frac{x}{a}} \\ & \mathrm{I}=a\left[\frac{x}{a} \tan ^{-1} \sqrt{\frac{x}{a}}-\sqrt{\frac{x}{a}}+\tan ^{-1} \sqrt{\frac{x}{a}}\right]+\mathrm{C} \\ & \mathrm{I}=(a+x) \tan ^{-1} \sqrt{\frac{x}{a}}-\sqrt{a x}+\mathrm{C}\end{aligned}$
Hence, the correct answer is option: 'A'.